Hi…i
There is a problem & I have solved it too however with one method things work out & with the other I am basically getting confused, could you please explain the facts
Ques] find the no., of two digit nos., where the product of the nos., is more than the sum of the nos.
Solution] we negate the statement such a*b< = a+b # if we are able to solve this then the
complement of this gives the answer => a*b-a-b+1 < = 1 => a[b-1]-1[b-1] < =1
so, a =1 or b =1 or a=b-=2 or b= 0 thus from 11 to 19 we get 9 nos{when a =1}&when b=1 ,from 21 to 91 we get 8 nos& when b=0 we get 10 to90 {9 nos}& 22 { 1 number}
so the no of 2 digits is 90-27=63 [ adding all we get 27]
however
if we do a*b> a+b then we get a[b-1]-1[b-1]>1……=>[a-1][b-1] >1
WHY IS THIS APPARENT CONTRADICTION…..could you ALSO give me the name of chapter I should read to get these concepts more cleared
REGARDS
sujoy
There is a problem & I have solved it too however with one method things work out & with the other I am basically getting confused, could you please explain the facts
Ques] find the no., of two digit nos., where the product of the nos., is more than the sum of the nos.
Solution] we negate the statement such a*b< = a+b # if we are able to solve this then the
complement of this gives the answer => a*b-a-b+1 < = 1 => a[b-1]-1[b-1] < =1
so, a =1 or b =1 or a=b-=2 or b= 0 thus from 11 to 19 we get 9 nos{when a =1}&when b=1 ,from 21 to 91 we get 8 nos& when b=0 we get 10 to90 {9 nos}& 22 { 1 number}
so the no of 2 digits is 90-27=63 [ adding all we get 27]
however
if we do a*b> a+b then we get a[b-1]-1[b-1]>1……=>[a-1][b-1] >1
WHY IS THIS APPARENT CONTRADICTION…..could you ALSO give me the name of chapter I should read to get these concepts more cleared
REGARDS
sujoy